Indicate the direction of polarity of each of the covalent bonds by placing the appropriate delta notation next to each end of the bond. the first bond is between c and o. the second bond is between o and c l. the third bond is between o and f. the fourth bond is between c and n. the fifth bond is between c l and c. the sixth bond is between s and h. the seventh bond is between s and c l.

(δ+) C → O (δ-)

(δ-) O ← Cl (δ+)

(δ+) O → F (δ-)

(δ+) C → N (δ-)

(δ-) Cl ← C (δ+)

(δ-) S ← H (δ+)

(δ+) S → Cl (δ-)

Explanation:

The polarity of a bond depends on the difference of electronegativity (EN) between the atoms that form it. The dipole moment (μ) points to the atom with the highest electronegativity, which is the one with a negative charge density (δ-), while the other atom has a positive charge density (δ+).

The first bond is between C and O.

EN(C) = 2.5; EN (O) = 3.5

(δ+) C → O (δ-)

The second bond is between O and Cl.

EN(O) = 3.5 ; EN(Cl) = 3.0

(δ-) O ← Cl (δ+)

The third bond is between O and F.

EN(O) = 3.5; EN(F) = 4.0

(δ+) O → F (δ-)

The fourth bond is between C and N.

EN(C) = 2.5; EN(N) = 3.0

(δ+) C → N (δ-)

The fifth bond is between Cl and C.

EN(Cl) = 3.0; EN(C) = 2.5

(δ-) Cl ← C (δ+)

The sixth bond is between S and H.

EN(S) = 2.5; EN(H) = 2.1

(δ-) S ← H (δ+)

The seventh bond is between S and Cl.

EN(S) = 2.5; EN(Cl) = 3.0

(δ+) S → Cl (δ-)

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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